model 3 twice, thrice, one third etc. of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 9 EXERCISES

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The following question based on average topic of quantitative aptitude

Questions : Of the three numbers, the first is 3 times the second and the third is 5 times the first. If the average of the three numbers is 57, the difference between the largest and the smallest number is

(a) 126

(b) 135

(c) 9

(d) 18

The correct answers to the above question in:

Answer: (a)

Let second number be x

∴ The first number = 3x

and the third number = 15x

Now, x + 3x + 15x = 3 × 57

⇒ 19x = 3 × 57

⇒ x= ${3×57}/19$= 9

∴ Required difference

= 15x – x = 14x

= 14 × 9 = 126

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

a = 3, b = 5, x = 57

First number = $\text"3ab"/ \text"1+b+ab"$x

= ${3×3×5}/{1+3+15}$× 57

= $45/19$ × 57 = 135

Second number = $\text"3ab"/ \text"1+b+ab"$x

= ${3×5}/{19}$× 57

= $15/19$ × 57 = 45

Third number = $3/ \text"1+b+ab"$x

= $3/19$×57 = 9

Required result = 135 – 9 = 126

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Read more twice thrice one third Based Quantitative Aptitude Questions and Answers

Question : 1

If the arithmetic mean of 3a and 4b is greater than 50, and a is twice b, then the smallest possible integer value of a is

a) 21

b) 19

c) 20

d) 18

Answer: (a)

$\text"3a+4b"/2$ > 50

⇒ 3a + 4b > 100

⇒ 3a + ${4a}/2$ > 100 [Since a = 2b]

⇒ 3a + 2a > 100

⇒ 5a > 100

⇒ a > 20

∴ Minimum value of a = 21

Question : 2

Of the three numbers, the first number is twice of the second and the second is thrice of the third number. If the average of these 3 numbers is 20, then the sum of the largest and smallest numbers is

a) 54

b) 60

c) 24

d) 42

Answer: (d)

Let the third number be x.

∴ Second number = 3x

and first number = 6x

∴ 6x + 3x + x = 3 × 20

⇒ 10x = 60 ⇒ x = 6

∴ Required sum = 6x + x = 7x = 7 × 6 = 42

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

Here, a = 2, b = 3, x = 20

Largest Number= $\text"3ab"/ \text"1+b+ab"$ x

= ${3×2×3}/{1+3+2×3}$×20

=$18/10$×20 = 36

Smallest Number = $3/\text"1+b+ab"$ ×x

= $3/{1+3+2×3}$×20

= $3/10$×20 = 6

Sum = 36 + 6 = 42

Question : 3

The average of first three numbers is thrice the fourth number. If the average of all the four numbers is 5, then find the fourth number.

a) 2

b) 4

c) 4.5

d) 5

Answer: (a)

Let the numbers be a, b, c,

${\text"a + b + c"}/3$ = 3d

⇒ a + b + c = 9d

Again, ${\text"a + b + c + d"}/4$ = 5

⇒ a + b + c + d = 20

⇒ 9d + d = 20

⇒ 10d = 20 ⇒ d = 2

Question : 4

Of the three numbers, the first is twice the second and the second is 3 times the third. If their average is 100, the largest of the three numbers is :

a) 180

b) 300

c) 120

d) 150

Answer: (a)

Let the third number be x.

∴ Second number = 3x

First number = 6x

∴ (x + 3x + 6x) = 100 × 3

⇒ 10x = 300

⇒ x = 30

∴ The largest number = 6x = 6 × 30 = 180

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

a = 2, b = 3, x = 100

Largest number = $\text"3ab"/ \text"1+b+ab"$x

= ${3×2×3}/{1+3+2×3}$×100

= ${18 × 100}/10$ = 180

Question : 5

Among three numbers, the first is twice the second and thrice the third. If the average of the three numbers is 49.5, then the difference between the first and the third number is

a) 39.5

b) 41.5

c) 54

d) 28

Answer: (c)

Let the second number be x.

∴ First number = 2x

∴ Third number =${2x}/3$

∴ 2x+x+${2x}/3$ = 49.5×3

⇒ 6x + 3x + 2x =49.5×9 = 445.5

⇒ 11x = 445.5

⇒ x = ${445.5}/11$ = 40.5

∴ Required difference

= 2x - ${2x}/3$ = ${4x}/3$

= ${4×40.5}/3$ = 54

Aliter : Using Rule 15,

From three numbers, first number is 'a’ times of 2nd number, 2nd number is 'b’ times of 3rd number and the average of all three numbers is x, then,

First number = $\text"3ab"/ \text"1+b+ab"$ x ; Second number = $\text"3b"/ \text"1+b+ab"$ x ; Third number = $\text"3b"/ \text"1+b+ab"$ x

Here, a = 2, b = $3/2$, x = 49.5

First Number = $\text"3ab"/ \text"1+b+ab"$ 5x

= ${3×2×{3/2}}/{1+{3/2}+2}×{3/2}$×49.5

= ${18/2}/{11/2}$×49.5

= ${18×49.5}/11$ = 18×4.5

= ${18×45}/10$ = 81

Third Number = $3/\text"1+b+ab"$ ×x

= $3/{1+{3/2}+2×{3/2}}$×49.5

= $3/{11/2}$ ×49 5.

= 6 ×4.5 = 27

Difference = 81 – 27 = 54

Question : 6

Of the three numbers whose average is 60, the first is one fourth of the sum of the others. The first number is :

a) 42

b) 45

c) 30

d) 36

Answer: (d)

x + y + z = 180

x =$1/4$(y+z)

⇒ 4x = y + z

⇒ 5x = 180,

∴ x = 36

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